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It’s fine to ask!
We don’t have a target date yet for that. Will be next year sometime, though. It takes awhile to put new courses together!
In our opinion, you can’t charge for not being able to diagnose something correctly/fully.
Most of the time you will be able to do an accurate diagnosis – especially as you gain experience. But occasionally you may end up not diagnosing it correctly and lose money on the parts. Sometimes this situation happens through no fault of your own – repairing one problem reveals another that you had no way of testing until the first fault was fixed, for example – but that’s one of the costs of doing business. It shouldn’t happen too often, but it’s important that your customers are given accurate quotes up front.
Do you charge a diagnostic fee?
Part of the challenge of this question is the funky way some of the lines are drawn. They have no meaning to the electrons! Everything comes down to is it a wire/switch or a load? Connections are not a load. Make sense?
Glad it helped!
In a series circuit, the current is the same throughout the circuit. So the current is calculated from the *total* resistance and the voltage supply. In this case, 37 ohms and 240vac (using I = E/R)
Once you have that current, you can then calculate the heat generated by an individual load using P = I2R
Hi John,
I’ll give you a (big) hint: the water that results from defrosting the evaporator is called “condensate”.
I know it’s referred to in the last video in that unit.
You are correct!
Great attention to detail 🙂
Hello!
You have to make a couple of calculations to get the final answer. Let me help step you through this.
The problem gives you the voltage supply (L1 and L2, so 240vac) and the total resistance for the circuit. (Total resistance for a series circuit is just the sum of the individual resistances, so in this case, 5 + 32 = 37 ohms.)
Look at the Ohm’s Law chart. Do you see a way to calculate the circuit current, I, if you know the total resistance and voltage for the circuit?
Hi Steve,
There isn’t, but the Final exam is a bit more of a review than the midterm is. We’ve found that folks who did well on the quizzes and module exams generally do well on the final if they’ve taken the time to look back through those quizzes/exams to refresh their memories.
It’s pretty straightforward- you might be overthinking it.
If you know the circuit current, you can find the voltage drop across each resistance using a simple Ohm’s Law calculation.
Exactly! That’s the case for a series circuit. Just wanted to make sure you were clear on that.
E=IxR can be used in a few different ways, but as far as calculating voltage drop across an individual load, then the way you wrote it in detail (the second one) is correct.
One clarification: you said “I (current flow through load)”. If you had several loads in a series circuit, would the current change at all? Would it be different for different loads, or the same throughout the circuit?
Hi Steve,
3c) voltage drop is the product of current flowing through a resistance (load). Does that make you think of an Ohm’s Law equation that you could use?
5b) re-watch the video at the end of unit 6.
let me know if you have any other questions!
August 9, 2016 at 2:05 pm in reply to: Unit #5 Basic Electricity: Series and Parallel Circuits #10643Way to pay attention! 🙂
You are correct – that is a mistake. Current is indeed inversely proportional to resistance. Good eye! Thanks for letting us know so we can correct that.
Good! I’m glad that you asked the question. We get some students who hesitate to ask questions here, which is a shame.
Yep! And Figure 2 is shows the testing that can tell you which line is out of commission.
Does that all seem clearer to you now? Any more questions?
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